Find the equations of the tangents to the circle `x^2+y^2-6x+4y=12` which are parallel to the straight line `4x+3y+5=0`

Explanation

Given circle is `x^(2)+y^(2)-6x+4y-12=0`
Centre of the circle is `(3,-2)` and radius is
`sqrt((-3)^(2)+(2)^(2)-(-12))=5`
Given that tangents are parallel to the line `4x+3y+5=0`.
Therefore, the equations of tangents take the form `4x+3y+c=0`.
Distance of centre of the circle from the tangent line is radius.
`:. |(4(3)+3(-2)+k)/(sqrt(16+9))|=5`
`implies 6+k=+-25`
`implies k=19` and `-31`
Therefore, the equations of required tangents are `4x+3y+19=0` and `4x+3y-31=0`
Alternative method `:`
Given circle in center-radius form form is `(x-3)^(2)+(y+2)^(2)=5^(2)`
Tangents are parallel to the line `4x+3y+5=0` .
`:. ` Slope of tangents `=-(4)/(3)`
Hence, using the equation of tangents
`y+f=m(x+g)+-sqrt(m^(2)+1)sqrt(g^(2)+f^(2)-c)`,
required equations of tangents are obtained as
`y+2=-(4)/(3)(x-3)+-5sqrt((-(4)/(3))^(2)+1)`
or `4x+3y+19=0` and `4x+3y-31=0`